Find the area of the circle in which a square of area $64{\text{c}}{{\text{m}}^2}$ is inscribed. (Use $\pi = 3.14$).

Hint: Here as we are given that square is inscribed in a circle. Therefore we can say that the diagonal of the square will be equal to the diameter of the circle and hence we will get the radius also of the circle. Then we can easily find the area of the circle by applying the formula $\pi {R^2}$.

Complete Step by Step Solution:
Here we are given that square has the area of $64{\text{c}}{{\text{m}}^2}$
As we know that area of the square is equal to ${\left( {{\text{side}}} \right)^2}$
Hence we can say that
$
  {\left( {{\text{side}}} \right)^2} = 64 \\
  {\text{side's length}} = \sqrt {64} = 8{\text{cm}} \\
 $
Therefore we can easily draw the given statement in the form of figure and we will get the diagram as:

So here we can see that $BD$ can also be called as the diameter of the circle as we know that if angle $BCD$ is right angled then the side opposite to the right angle will be equal to the diameter of the circle.
Hence we can say that $BD = {\text{diameter}} = 2r$ and here $r = {\text{radius}}$
Therefore we can say that:
$2r = {\text{diagonal of the square}}$
We have written this because from the figure it is very clear that diagonal of the square is equal to the diameter of the circle.
We know that:
Diagonal of the square is given by $\sqrt 2 \left( {{\text{length of side}}} \right) = \sqrt 2 \left( 8 \right) = 8\sqrt 2 {\text{cm}}$
Hence we can equate diameter which is twice the radius with this length of the diagonal and we will get:
$
  2r = 8\sqrt 2 \\
  r = 4\sqrt 2 {\text{cm}} \\
 $
Hence we have got the radius and now we can calculate the area of the circle by using the formula $\pi {R^2}$

Area$ = \pi {R^2} = \pi {\left( {4\sqrt 2 } \right)^2} = 32\pi {\text{ c}}{{\text{m}}^2}$$32\left( {3.14} \right) = 100.48{\text{c}}{{\text{m}}^2}$.

Note:
Here if the student is given that circle is inscribed in the square then we need to again draw the figure and we will see that here diameter of the circle will be equal to the side of the square and hence we can again find the required result.